Problem: Find the smallest positive integer $b$ for which $x^2+bx+2008$ factors into a product of two binomials, each having integer coefficients.
Solution: The question implies that we can factor the given quadratic as \begin{align*}
x^2+bx+2008 &= (x+r)(x+s)\\
& = x^2+(r+s)x+rs, \end{align*} where $r$ and $s$ are integers.  Since both $b$ and 2008 are positive, it is clear that $r$ and $s$ must also be positive.  By multiplying out the right-hand side as shown, we see that we must have $rs=2008$, which has prime factorization $2008=2\cdot 2\cdot 2\cdot 251$.  Recall that we are trying to minimize $b=r+s$.  The best we can do is to let $r=251$ and $s=8$, leading to $b=251+8=\boxed{259}$.